 So we've been talking about lots of different types of processes that we can now calculate various thermodynamic quantities for. For example, when we talk about isothermal processes, those are processes which don't experience a change in temperature. The change in temperature is zero. We've also talked about isobaric processes which don't change their pressure or isochoric processes, which don't have a change in volume. So there's one additional type of process we haven't talked about yet that turns out to be pretty important. And that's a process where the heat transfer in a process is equal to zero. So notice I don't put a delta in front of this because there's no q associated with the system that's changing. The q is the amount of heat flowing into or out of the system depending on the sign. So if there's no heat flow into or out of the system, then q is equal to zero. And we call that process adiabatic. So an adiabatic process is one where we don't allow heat to enter the system. If we have a system where heat is flowing in or perhaps in a different system heat is flowing out, that would be not adiabatic. In order to have an adiabatic system, what we need to do is take our system and somehow insulate it from the environment. So if I wrap it in an insulating layer, examples you might be familiar with would be a well insulated thermos that doesn't allow your hot coffee to cool down over the course of a day. Or in the chemistry department a doer flask that doesn't allow heat to enter and raise the temperature of the cold liquid that you've got inside the flask. If you've got a very well insulated box, then no heat flows from outside of the system to inside of the system. And any process taking place in the system will be an adiabatic process. So if we imagine what an adiabatic process would look like, let's say we want to talk about an adiabatic expansion or an adiabatic compression of a gas. So some sort of adiabatic change in the volume. I need to have, so before when we talked about volume changes, we've typically used a diagram like a piston. So we have some P1, V1 inside a piston and as the gas expands, let's say to some lower P2 and some larger V2, the piston moves upward to allow it to increase the volume. So same thing here except we just need to have a very well insulated piston do that gas expansion inside of a doer flask or inside of a very well insulated container of some sort. So that process is what we mean by an adiabatic process. If we think about what's going on in that adiabatic process, by definition if it's adiabatic, there's no heat transfer in the process. So the heat transfer is zero. So dU must be equal to dW. We know for a gas expansion, let's say in particular that we're talking about an expansion, we know for a gas expansion, the system is doing work on the surroundings. The lift of the box is being pushed up by the system. The system requires some energy to do that work. It costs the system some energy and we represent that with a negative sign on the PV work. So that means also necessarily the internal energy is also negative. So the internal energy of the system is going to decrease because if we don't allow heat to flow into the system to pay for that PV work of expanding the gas, the only place that energy can come from is the internal energy that the gas already contains. So in order to perform the expansion, the internal energy is going to decrease and what that means is the temperature of the gas is also going to decrease. So what this means is when we do an adiabatic expansion of a gas, its temperature will drop. And conversely, as we'll see in just a second without working through all these details again, if I were to do an adiabatic compression, if I compress a well insulated gas, its temperature will increase. And that's actually something you probably have some experience with when you let a gas expand, let's say out of a pumped up bicycle tire or out of a can of compressed air that used to clean your keyboard. The gas coming out of the nozzle is typically colder than room temperature. So let's say we're curious about exactly how much the temperature of that gas decreases. So let's do the same thing a little more quantitatively now. So still talking about an adiabatic volume change, we had gotten to the point of saying du must be equal to dw because it's adiabatic and dq must be zero. We know that the internal energy change is related to the temperature change, moles times the molar heat capacity times the change in temperature. And on the right side of the equation, the work is equal to minus p external dv if we do it irreversibly or minus p, the system pressure if we do it reversibly. So let's say we're doing a reversible adiabatic volume change so that I can write p instead of p external. And let's in addition say that we're doing this for an ideal gas. That will allow me to write this pressure as nrt over v and on the left side I've still got ncv times dt. So that's starting to look good. What I would normally do at this point in order to relate how much the temperature changes to how much the volume changes for a finite amount of changes, I'd throw an integral sign in front of both of these. But we're not quite ready to do that yet because this integral on the right hand side I'm integrating with respect to v and it's fine that I've got a v here but I don't like the temperature that shows up here. Likewise I'd rather have the temperature over on this side where I'm talking about the volume change. So if I just divide each side of this expression by temperature, so I've got ncv bar 1 over t dt is equal to minus nr 1 over v dv. Then I'm ready to do a couple things. I can put an integral sign on both sides of this equation, I can also notice that I've got an n on the left and an n on the right so I can cancel those two ends. And when I do this integral I find that on the left cv bar log of final conditions over initial conditions log of t2 over t1 must be equal to on the right side, don't forget the negative sign r log final conditions over initial conditions but now we're talking about the volume v2 over v1. So our goal remember was to ask ourselves how much does the temperature change when I do this adiabatic expansion or compression of a gas so I want to rearrange this equation and find solve for the amount of the temperature change as a function of the volume change. So let's write log of t2 over t1 if I move the heat capacity over to the right I've got minus r over cv molar cv log of v2 over v1. This term looks like a number multiplying this logarithm. So a constant times a log is equal to the log raised to that constant so I'll rewrite this, let's leave the negative sign out front, negative log of v2 over v1 raised to the r over cv bar power. And I'll use one more log to say the negative of a log is equal to the log of the inverse so instead of v2 over v1 I've actually got log of v1 over v2 if I use a positive sign. Log of t2 over t1 is equal to log of v1 over v2 raised to that power r divided by cv and now I've got log on the left, log on the right which I can remove both those logs essentially exponentiating both sides of that equation and then that will be our final result. So that one we'll put in a box and we'll use that again soon. What that tells us is the connection between the temperature change and the volume change for this adiabatic expansion or compression. If we know the heat capacity of the gas and if we know the gas constant then let's say if I expand the gas to twice its initial volume if I do an expansion so that v2 is larger than v1, v1 over v2 is a number less than one and that means that the result on the right hand side will be the temperature t2 will be colder than the temperature t1. On the other hand if I do an adiabatic compression and v2 is larger than v1 then we'll work the other way around and the final temperature will be larger than the initial temperature. So we'll use this equation to perform a numerical example to see how that works but at the moment what we've got is at least a qualitative understanding of why it is that when I do an adiabatic expansion the temperature drops and when I do an adiabatic compression the temperature increases. So the next step will be just plug some numbers into this and see how it works.